Further Maths Flows in networks Practice Questions

Free Further Maths Flows in networks practice questions with full step-by-step worked solutions. Covers flows, max-flow-min-cut, networks, value-of-a-flow. Practise exam-style problems and check your method.

flowsmax-flow-min-cutnetworksvalue-of-a-flowcutscapacity-of-a-cut
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The directed network NN has source SS and sink TT. The arcs and their capacities are {SA:4, SB:3, AB:2, AT:3, BT:5}\{S\to A:4,\ S\to B:3,\ A\to B:2,\ A\to T:3,\ B\to T:5\}. Find the value of the maximum flow from SS to TT.
Show worked solution

Worked solution

  1. Write down the network as data

    N: {SA:4, SB:3, AB:2, AT:3, BT:5}N:\ \{S\to A:4,\ S\to B:3,\ A\to B:2,\ A\to T:3,\ B\to T:5\}

    The source, the sink, the arcs and their capacities describe the whole network.

  2. Add the capacities of the arcs leaving SS

    vc(S,v)=4+3=7\sum_{v}c(S,v)=4+3=7

    These are the arcs pointing away from SS.

  3. State the value of the maximum flow

    maxval(f)=7\max\operatorname{val}(f)=7

    No larger flow can be sent from the source to the sink.

Answer
maxval(f)=7\max\operatorname{val}(f)=7
Question 2
2 markseasy
The directed network NN has source SS and sink TT. The arcs and their capacities are {SA:4, SB:3, AB:2, AT:3, BT:5}\{S\to A:4,\ S\to B:3,\ A\to B:2,\ A\to T:3,\ B\to T:5\}. By the maximum-flow minimum-cut theorem, which of the following is the value of the maximum flow from SS to TT?
Show worked solution

Worked solution

  1. Write down the network as data

    N: {SA:4, SB:3, AB:2, AT:3, BT:5}N:\ \{S\to A:4,\ S\to B:3,\ A\to B:2,\ A\to T:3,\ B\to T:5\}

    The source, the sink, the arcs and their capacities describe the whole network.

  2. Add the capacities of the arcs leaving SS

    vc(S,v)=4+3=7\sum_{v}c(S,v)=4+3=7

    These are the arcs pointing away from SS.

  3. Select the value of the maximum flow

    maxval(f)=7\max\operatorname{val}(f)=7

    This value is achieved and cannot be exceeded.

Answer
maxval(f)=7\max\operatorname{val}(f)=7
Question 3
4 marksintermediate
The directed network NN has source SS and sink TT. The arcs and their capacities are {SA:3, SB:6, AB:4, AC:5, BC:2, BD:5, CT:4, DT:6, CD:2}\{S\to A:3,\ S\to B:6,\ A\to B:4,\ A\to C:5,\ B\to C:2,\ B\to D:5,\ C\to T:4,\ D\to T:6,\ C\to D:2\}. The flow ff assigns arc values {SA:3, SB:5, AB:0, AC:3, BC:0, BD:5, CT:3, DT:5, CD:0}\{S\to A:3,\ S\to B:5,\ A\to B:0,\ A\to C:3,\ B\to C:0,\ B\to D:5,\ C\to T:3,\ D\to T:5,\ C\to D:0\}. Which of the following is a flow-augmenting path from SS to TT, together with the increase it gives?
Show worked solution

Worked solution

  1. Write down the network as data

    N: {SA:3, SB:6, AB:4, AC:5, BC:2, BD:5, CT:4, DT:6, CD:2}N:\ \{S\to A:3,\ S\to B:6,\ A\to B:4,\ A\to C:5,\ B\to C:2,\ B\to D:5,\ C\to T:4,\ D\to T:6,\ C\to D:2\}

    The source, the sink, the arcs and their capacities describe the whole network.

  2. Write down the given flow

    f: {SA:3, SB:5, AB:0, AC:3, BC:0, BD:5, CT:3, DT:5, CD:0}f:\ \{S\to A:3,\ S\to B:5,\ A\to B:0,\ A\to C:3,\ B\to C:0,\ B\to D:5,\ C\to T:3,\ D\to T:5,\ C\to D:0\}

    Each arc carries the amount of flow shown.

  3. Find the spare capacity on the forward arc SBS\to B

    c(S,B)f(S,B)=65=1c(S,B)-f(S,B)=6-5=1

    A forward arc can take up to its spare capacity.

  4. Find the spare capacity on the forward arc BCB\to C

    c(B,C)f(B,C)=20=2c(B,C)-f(B,C)=2-0=2

    A forward arc can take up to its spare capacity.

  5. Find the spare capacity on the forward arc CDC\to D

    c(C,D)f(C,D)=20=2c(C,D)-f(C,D)=2-0=2

    A forward arc can take up to its spare capacity.

  6. Find the spare capacity on the forward arc DTD\to T

    c(D,T)f(D,T)=65=1c(D,T)-f(D,T)=6-5=1

    A forward arc can take up to its spare capacity.

  7. Select the flow-augmenting path

    δ=1\delta=1

    Every arc on this path has spare capacity, so the flow can increase.

Answer
δ=1\delta=1
Question 4
6 markshard
The directed network NN has source SS and sink TT. The arcs and their capacities are {SA:3, SB:6, AB:4, AC:5, BC:2, BD:5, CT:4, DT:6, CD:2}\{S\to A:3,\ S\to B:6,\ A\to B:4,\ A\to C:5,\ B\to C:2,\ B\to D:5,\ C\to T:4,\ D\to T:6,\ C\to D:2\}. Which of the following is a minimum cut of NN, together with its capacity?
Show worked solution

Worked solution

  1. Write down the network as data

    N: {SA:3, SB:6, AB:4, AC:5, BC:2, BD:5, CT:4, DT:6, CD:2}N:\ \{S\to A:3,\ S\to B:6,\ A\to B:4,\ A\to C:5,\ B\to C:2,\ B\to D:5,\ C\to T:4,\ D\to T:6,\ C\to D:2\}

    The source, the sink, the arcs and their capacities describe the whole network.

  2. Add the capacities of the arcs leaving SS

    vc(S,v)=3+6=9\sum_{v}c(S,v)=3+6=9

    These are the arcs pointing away from SS.

  3. Identify the arcs crossing the cut from the source side

    SA, SBS\to A,\ S\to B

    Only arcs from the source side to the sink side are counted.

  4. Add the capacities of those arcs

    C=3+6=9C=3+6=9

    Arcs pointing back from the sink side are ignored.

  5. Confirm the flow reaches this cut value

    maxval(f)=9\max\operatorname{val}(f)=9

    A cut whose capacity equals the maximum flow is a minimum cut.

  6. Recall the value of a flow

    val(f)=vf(S,v)vf(v,S)\operatorname{val}(f)=\sum_{v}f(S,v)-\sum_{v}f(v,S)

    The value of a flow is the net flow out of the source.

  7. Recall flow conservation at an intermediate vertex

    uf(u,x)=wf(x,w)\sum_{u}f(u,x)=\sum_{w}f(x,w)

    At every vertex except the source and the sink, flow in equals flow out.

  8. Recall the capacity constraint on each arc

    0f(u,v)c(u,v)0\le f(u,v)\le c(u,v)

    No arc may carry more flow than its capacity.

  9. Recall the definition of a cut

    a cut splits the vertices into a source side and a sink side\text{a cut splits the vertices into a source side and a sink side}

    The source lies on one side of the cut and the sink on the other.

  10. Recall the capacity of a cut

    C=uS-side, vT-sidec(u,v)C=\sum_{u\in S\text{-side},\ v\in T\text{-side}}c(u,v)

    Only arcs crossing from the source side to the sink side are counted.

  11. Select the minimum cut

    minC=9\min C=9

    Its capacity equals the maximum flow, which proves it is minimal.

Answer
minC=9\min C=9
Question 5
9 markschallenging
The directed network NN has source SS and sink TT. The arcs and their capacities are {SA:5, SB:6, AC:4, AD:3, BD:5, CT:6, DT:4, BC:2}\{S\to A:5,\ S\to B:6,\ A\to C:4,\ A\to D:3,\ B\to D:5,\ C\to T:6,\ D\to T:4,\ B\to C:2\}. The flow ff assigns arc values {SA:4, SB:4, AC:4, AD:0, BD:4, CT:4, DT:4, BC:0}\{S\to A:4,\ S\to B:4,\ A\to C:4,\ A\to D:0,\ B\to D:4,\ C\to T:4,\ D\to T:4,\ B\to C:0\}. Find the value of the flow ff.
Show worked solution

Worked solution

  1. Write down the network as data

    N: {SA:5, SB:6, AC:4, AD:3, BD:5, CT:6, DT:4, BC:2}N:\ \{S\to A:5,\ S\to B:6,\ A\to C:4,\ A\to D:3,\ B\to D:5,\ C\to T:6,\ D\to T:4,\ B\to C:2\}

    The source, the sink, the arcs and their capacities describe the whole network.

  2. Write down the given flow

    f: {SA:4, SB:4, AC:4, AD:0, BD:4, CT:4, DT:4, BC:0}f:\ \{S\to A:4,\ S\to B:4,\ A\to C:4,\ A\to D:0,\ B\to D:4,\ C\to T:4,\ D\to T:4,\ B\to C:0\}

    Each arc carries the amount of flow shown.

  3. Add the flow along the arcs leaving the source SS

    val(f)=4+4=8\operatorname{val}(f)=4+4=8

    The value of a flow is the net flow out of the source.

  4. Recall the value of a flow

    val(f)=vf(S,v)vf(v,S)\operatorname{val}(f)=\sum_{v}f(S,v)-\sum_{v}f(v,S)

    The value of a flow is the net flow out of the source.

  5. Recall flow conservation at an intermediate vertex

    uf(u,x)=wf(x,w)\sum_{u}f(u,x)=\sum_{w}f(x,w)

    At every vertex except the source and the sink, flow in equals flow out.

  6. Recall the capacity constraint on each arc

    0f(u,v)c(u,v)0\le f(u,v)\le c(u,v)

    No arc may carry more flow than its capacity.

  7. Recall the definition of a cut

    a cut splits the vertices into a source side and a sink side\text{a cut splits the vertices into a source side and a sink side}

    The source lies on one side of the cut and the sink on the other.

  8. Recall the capacity of a cut

    C=uS-side, vT-sidec(u,v)C=\sum_{u\in S\text{-side},\ v\in T\text{-side}}c(u,v)

    Only arcs crossing from the source side to the sink side are counted.

  9. Recall the maximum-flow minimum-cut theorem

    maxval(f)=minC\max\operatorname{val}(f)=\min C

    The value of a maximum flow equals the capacity of a minimum cut.

  10. Recall the weak-duality bound

    val(f)C\operatorname{val}(f)\le C

    Every flow value is at most the capacity of every cut.

  11. Recall how a cut proves optimality

    val(f)=C  f is maximal\operatorname{val}(f)=C\ \Rightarrow\ f\text{ is maximal}

    If a flow and a cut have the same size, neither can be improved.

  12. Recall a saturated arc

    f(u,v)=c(u,v)f(u,v)=c(u,v)

    A saturated arc is full to its capacity.

  13. Recall the spare capacity of an arc

    spare=c(u,v)f(u,v)\text{spare}=c(u,v)-f(u,v)

    This is how much more flow the arc could still carry.

  14. Recall a flow-augmenting path

    a path from S to T with spare capacity throughout\text{a path from }S\text{ to }T\text{ with spare capacity throughout}

    Along such a path the value of the flow can be increased.

  15. Recall forward and backward arcs on an augmenting path

    forward: cf,backward: f\text{forward: }c-f,\qquad \text{backward: }f

    Forward arcs use spare capacity; backward arcs release existing flow.

  16. State the value of the flow

    val(f)=8\operatorname{val}(f)=8

    This is the net flow out of the source.

Answer
val(f)=8\operatorname{val}(f)=8

Unlock 65 more Flows in networks questions

Create a free account to work through every Further Maths Flows in networks question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Flows in networks practice

Related Decision Maths topics